It is a conjectured that $$\sum_{r=1}^n (r+1)2^r = (a+bn)2^n$$ where $a$ and $b$ are constants. If the conjecture is true, verify that $a=0$ and $b=2$ and prove the result by induction.
I have no idea how I would go about verifying the values for $a$ and $b$, so would someone be able to help me get started on that.
I have attempted the proof by induction, where my inductive step doesn't seem quite right to me. My workings are below is there anything wrong with it?
If $n=k$ $$ \sum_{r=1}^{k}(r+1)2^r = 2^k(2k)$$ If $n=k+1$ \begin{align*} \sum_{r=1}^{k+1}(r+1)2^r &= \sum_{r=1}^{k}(r+1)2^r + 2^{k+1}(k+2)\\ {}&=2^k(2k) + 2^{k+1}(k+2)\\ &=2^k(2k+2(k+2))\\ &=2^k(4k+4)\\ &=2^{k+1}(2(k+1))\\ &=2^n(2n) \end{align*}
The case $n=0$ gives $0=a$; the case $n=1$ gives $4=(a+b)2\implies b=2$. Now prove the sum is $n2^{n+1}$ by induction. I don't know why you have misgivings; your inductive step is fine. (We needn't check the base step, since the calculations above already did.)